Question

2.7 A 3400-lb car is traveling in third gear (overall gear reduction ratio of 2.5 to 1) on a level road at its top speed of 130 mi/h. The air density is 0.00206 slugs/ft3 . The car has a frontal area of 19.8 ft2 , a drag coefficient of 0.28, a wheel radius of 12.6 inches, a drive axle slippage of 3%, and a drivetrain efficiency of 88%. At this vehicle speed, what torque is the engine producing and what is the engine speed (in revolutions per minute)?

Answer 1

1) The engine torque is approximately 134.33 N·m

2) The speed of the engine is approximately 4,469.15 revolutions per minute

Explanation:

1) The drag coefficient,
`c_d`, is given by the formula;

`c_d = (2 \cdot F_d)/(\rho \cdot u^2 \cdot A)`

Where;

`c_d` = 0.28

`F_d` = The drag force

ρ = The fluid density = 0.00206 slugs/ft³ = 1.06168037 kg/m³

u = The object's flow speed = 130 mi/h = 58.1152 m/s

A = The frontal area = 19.8 ft² = 1.83948 m²

`F_d = (c_d \cdot \rho \cdot u^2 \cdot A )/(2)`

∴
`F_d` = (0.28 × 1.06168037 × (58.1152)² × 1.83948)/2 ≈ 923.4 N

We have;

`F_d = (M_e \cdot \varepsilon _0 \cdot \eta _d )/(r)`

Where;

`M_e` = The engine torque

ε₀ = The overall gear reduction ratio = 2.5

`\eta _d` = The drivetrain efficiency = 0.88

r = The wheel radius = 12.6 inches = 0.32004 meters

`\therefore M_e = (F_d \cdot r )/( \varepsilon _0 \cdot \eta _d)`

`\therefore M_e = (F_d \cdot r )/( \varepsilon _0 \cdot \eta _d) \approx (923.4 * 0.32004 )/( 2.5 * 0.88) \approx 134.33 \ N\cdot m`

The engine torque =
`M_e` ≈ 134.33 N·m

The engine torque ≈ 134.33 N·m

2) The speed of the engine,
`n_e`, is obtained from the following formula;

`v = (2 \cdot \pi \cdot r \cdot n_e \cdot (1 - i))/(\varepsilon _0)`

Where;

v = The vehicle's speed = 130 mi/h = 58.1152 m/s

r = The wheel radius = 12.6 inches = 0.32004 meters

i = The drive axle slippage = 3% = 3/100 = 0.03

ε₀ = The overall gear reduction ratio = 2.5

`\therefore n_e = (v * \varepsilon _0 )/(2 * \pi * r * (1 - i)) = (58.1152 *2.5 )/(2 * \pi * 0.32004 * (1 - 0.03)) \approx 74.486 \ rev /second`

The speed of the engine in revolution per minute = 60 seconds/minute × 74.486 rev/second ≈ 4,469.15 revolutions per minute

The speed of the engine ≈ 4,469.15 revolutions per minute.